I need to write a bash script that takes service name as a parameter and display only comment that is after hash symbol in /etc/services but I have no idea how to cut only the comment part.
The ,,it's working solution'' for me is to just:
grep "^$1" /etc/services | awk '{print $3,$4 ...
but I don't think this is a good one
I'm searching for something like:
[find the service] -> print only the part from # till the end of the line
I'm still learning so any solution with explanation or just a hint will be very helpful for me.
Chances are this is what you're looking for:
awk -v svc="$1" '($1==svc) && sub(/[^#]+#/,"")' /etc/services
but without sample input/output it's a guess.
The above will work using any awk in any shell on every Unix box.
Try this:
SERVICE_NAME=linuxconf; grep -Po "^$SERVICE_NAME.*# \K.*$" /etc/services
-P tells grep to use perl regex.
-o trims the output so that it only includes the regex match.
\K tells the regex engine to exclude previously matched part of the string from the match, i.e. only the part after \K will be present in the final match.
Related
Let's say I have a line looking like this:
/Users/random/354765478/Tests/StoreTests/Base64Tests.swift
In this example, I would like to get the result:
Tests/StoreTests/Base64Tests.swift
How can I do if I want to get everything before the first pattern match (either Sources or Tests) using sed or awk?
I am using sed 's/^.*\(Tests.*\).*$/\1/' right now but it's falling:
echo '/Users/random/354765478/Tests/StoreTests/Base64Tests.swift' | sed 's/^.*\(Tests\)/\1/'
Tests.swift
Here's another example using Sources (which seems to work):
echo '/Users/random/741672469/Sources/Store/StoreDataSource.swift' | sed 's/^.*\(Sources\)/\1/'
Sources/Store/StoreDataSource.swift
I would like to get everything before the first, and not the last Sources or Tests pattern match.
Any help would be appreciated!
How can I do if I want to get everything before the first pattern match (either Sources or Tests).
Easier to use a grep -o here:
grep -Eo '(Sources|Tests)/.*' file
Tests/StoreTests/Base64Tests.swift
Sources/Store/StoreDataSource.swift
# where input file is
cat file
/Users/random/354765478/Tests/StoreTests/Base64Tests.swift
/Users/random/741672469/Sources/Store/StoreDataSource.swift
Breakdown:
Regex pattern (Sources|Tests)/.* match any text that starts with Sources/ or Tests/ until end of the line.
-E: enables extended regex mode
-o: prints only matched text instead of full line
Alternatively you may use this awk as well:
awk 'match($0, /(Sources|Tests)\/.*/) {
print substr($0, RSTART)
}' file
Tests/StoreTests/Base64Tests.swift
Sources/Store/StoreDataSource.swift
Or this sed:
sed -E 's~.*/((Sources|Tests)/.*)~\1~' file
Tests/StoreTests/Base64Tests.swift
Sources/Store/StoreDataSource.swift
With your shown samples please try following GNU grep. This will look for very first match of /Sources OR /Tests and then print values from these strings to till end of the value.
grep -oP '^.*?\/\K(Sources|Tests)\/.*' Input_file
Using sed
$ sed -E 's~([^/]*/)+((Tests|Sources).*)~\2~' input_file
Tests/StoreTests/Base64Tests.swift
would like to get everything before the first, and not the last
Sources or Tests pattern match.
First thing is to understand reason of that, you are using
sed 's/^.*\(Tests.*\).*$/\1/'
observe that * is greedy, i.e. it will match as much as possible, therefore it will always pick last Tests, if it would be non-greedy it would find first Tests but sed does not support this, if you are using linux there is good chance that you have perl command which does support that, let file.txt content be
/Users/random/354765478/Tests/StoreTests/Base64Tests.swift
then
perl -p -e 's/^.*?(Tests.*)$/\1/' file.txt
gives output
Tests/StoreTests/Base64Tests.swift
Explanation: -p -e means engage sed-like mode, alterations in regular expression made: brackets no longer require escapes, first .* (greedy) changed to .*? (non-greedy), last .* deleted as superfluous (observe that capturing group will always extended to end of line)
(tested in perl 5, version 30, subversion 0)
Have been stuck with this little puzzle. Thank you in advance for helping.
I have a directory path and would like print its path after match.
like
echo /Users/user/Documents/terraform-shared-infra/services/history_book_test | awk -F "terraform-|tfRepo-" '{print $(NF)}'
echo /Users/user/Documents/tfRepo-shared-infra/services/history_book_test | awk -F "terraform-|tfRepo-" '{print $(NF)}'
output:
shared-infra/services/history_book_test
shared-infra/services/history_book_test
When i try to add wildcard in terraform-* it doesn't work.
I would like to print path after match with terraform-* or tfRepo*.
Like:
services/history_book_test
services/history_book_test/../.. so on.
with sed:
echo /Users/user/Documents/terraform-shared-infra/services/history_book_test | sed 's|.*terraform.\([^/]*\)/.*|\1|'
shared-infra
Have tried different ways with awk and grep but no luck. Any leads or idea that I can try. Please.
Thank you.
You're confusing regular expressions with globbing patterns. Both have wildcards and look similar but have quite different meanings and uses. regexps are used by text processing tools like grep, sed, and awk to match text in input strings while globbing patterns are used by shells to match file/directory names. For example, foo* in a regexp means fo followed by zero or more additional os while foo* in a globbing pattern means foo followed by zero or more other characters (which in a regexp would be foo.*). So never just say "wildcard", say "regexp wildcard" or "globbing wildcard" for clarity.
This might be what you're trying to do, using a sed that has a -E arg to enable EREs, e.g. GNU or BSD sed:
$ sed -E 's:.*/(terraform|tfRepo)-[^/]*/::' file
services/history_book_test
services/history_book_test
or using any awk:
$ awk '{sub(".*/(terraform|tfRepo)-[^/]*/","")} 1' file
services/history_book_test
services/history_book_test
Regarding your attempt with sed sed 's|.*terraform.\([^/]*\)/.*|\1|' - if you're going to use a char other than / for the delimiters, don't use a char like | that's a regexp or backreference metachar as at best that obfuscates your code, pick some char that's always literal instead, e.g. :.
Each file's name starts with "input". One example of the files look like:
0.0005
lii_bk_new
traj_new.xyz
0
73001
146300
I want to delete the lines which only includes '0' and the expected output is:
0.0005
lii_bk_new
traj_new.xyz
73001
146300
I have tried with
sed -i 's/^0\n//g' input_*
and
grep -RiIl '^0\n' input_* | xargs sed -i 's/^0\n//g'
but neither works.
Please give some suggestions.
Could you please try changing your attempted code to following, run it on a single Input_file once.
sed 's/^0$//' Input_file
OR as per OP's comment to delete null lines:
sed 's/^0$//;/^$/d' Input_file
I have intentionally not put -i option here first test this in a single file of output looks good then only run with -i option on multiple files.
Also problem in your attempt was, you are putting \n in regex of sed which is default separator of line, we need to put $ in it to tell sed delete those lines which starts and ends with 0.
In case you want to take backup of files(considering that you have enough space available in your file system) you could use -i.bak option of sed too which will take backup of each file before editing(this isn't necessary but for safer side you have this option too).
$ sed '/^0$/d' file
0.0005
lii_bk_new
traj_new.xyz
73001
146300
In your regexp you were confusing \n (the literal LineFeed character which will not be present in the string sed is analyzing since sed reads one \n-separated line at a time) with $ (the end-of-string regexp metacharacter which represents end-of-line when the string being parsed is a line as is done with sed by default).
The other mistake in your script was replacing 0 with null in the matching line instead of just deleting the matching line.
Please give some suggestions.
I would use GNU awk -i inplace for that following way:
awk -i inplace '!/^0$/' input_*
This simply will preserve all lines which do not match ^0$ i.e. (start of line)0(end of line). If you want to know more about -i inplace I suggest reading this tutorial.
I went through several grep examples, but don't see how to do the following.
Say, i have a file with a line
! some test here and number -123.2345 text
i can get this line using
grep ! input.txt
but how do i get the number (possibly positive or negative) from this line and append it to the end of another file? Is it possible to apply grep to grep results?
If yes, then i could get the number via something like
grep -Eo "[0-9]{1,}|\-[0-9]{1,}"
p/s/ i am using OS-X
p/p/s/ i'm trying to fetch data from several files and put into a single file for later plotting.
The format with your commands would be:
grep ! input.txt | grep -Eo "[0-9]{1,}|\-[0-9]{1,}" >> output
To grep from grep we use the pipe operator | this lets us chain commands together. To append this output to a file we use the redirection operator >>.
However there are a couple of problems. You regexp is better written: grep -Eoe '-?[0-9.]+' this allows for the decimal and returns the single number instead of two and if you want lines that start with ! then grep ^! is better to avoid matches with lines what contain ! but don't start with it. Better to do:
grep '^!' input | grep -Eoe '-?[0-9.]+' >> output
perl -lne 'm/.*?([\d\.\-]+).*/g;print $1' your_file >>anotherfile_to_append
$foo="! some test here and number -123.2345 text"
$echo $foo | sed -e 's/[^0-9\.-]//g'
$-123.2345
Edit:-
for a file,
[ ]$ cat log
! some test here and number -123.2345 text
some blankline
some line without "the character" and with number 345.566
! again a number 34
[ ]$ sed -e '/^[^!]/d' -e 's/[^0-9.-]//g' log > op
[ ]$ cat op
-123.2345
34
Now lets see the toothpicks :) '/^[^!]/d' / start of pattern, ^ not (like multiply with false), [^!] anyline starting with ! and d delete. Second expression, [^0-9.-] not matching anything within 0 to 9, and . and -, (everything else) // replace with nothing (i.e. delete) and done :)
I am trying to search my rails directory using grep. I am looking for a specific word and I want to grep to print out the file name and line number.
Is there a grep flag that will do this for me? I have been trying to use a combination of -n and -l but these are either printing out the file names with no numbers or just dumping out a lot of text to the terminal which can't be easily read.
ex:
grep -ln "search" *
Do I need to pipe it to awk?
I think -l is too restrictive as it suppresses the output of -n. I would suggest -H (--with-filename): Print the filename for each match.
grep -Hn "search" *
If that gives too much output, try -o to only print the part that matches.
grep -nHo "search" *
grep -rin searchstring * | cut -d: -f1-2
This would say, search recursively (for the string searchstring in this example), ignoring case, and display line numbers. The output from that grep will look something like:
/path/to/result/file.name:100: Line in file where 'searchstring' is found.
Next we pipe that result to the cut command using colon : as our field delimiter and displaying fields 1 through 2.
When I don't need the line numbers I often use -f1 (just the filename and path), and then pipe the output to uniq, so that I only see each filename once:
grep -ir searchstring * | cut -d: -f1 | uniq
I like using:
grep -niro 'searchstring' <path>
But that's just because I always forget the other ways and I can't forget Robert de grep - niro for some reason :)
The comment from #ToreAurstad can be spelled grep -Horn 'search' ./, which is easier to remember.
grep -HEroine 'search' ./ could also work ;)
For the curious:
$ grep --help | grep -Ee '-[HEroine],'
-E, --extended-regexp PATTERNS are extended regular expressions
-e, --regexp=PATTERNS use PATTERNS for matching
-i, --ignore-case ignore case distinctions
-n, --line-number print line number with output lines
-H, --with-filename print file name with output lines
-o, --only-matching show only nonempty parts of lines that match
-r, --recursive like --directories=recurse
Here's how I used the upvoted answer to search a tree to find the fortran files containing a string:
find . -name "*.f" -exec grep -nHo the_string {} \;
Without the nHo, you learn only that some file, somewhere, matches the string.